Why were methylene chloride and hydrochloric acid used to obtain the product?
From “Exp 006
2:30pm (04/02/2006) Extraction: Added 5ml of 5% HCl and 5 ml methylenechloride to the round bottom flask containg the crude product and transferred it to a separatory funnel. Separated the bottom methylenechloride layer from the top aqueous layer. Washed the aquoues acid layer twice with 5mls of methylenechloride and collected the bottom methylenechloride layer each time. Combined the organic layers and dried it over anhydrous magnesium sulfate (MgSO4). Evaporated the dichloromethane using a rotovap and collected the extracted product. A TLC of the product was obtained (Shown above).”
Because the product is soluble in methylene chloride , the methylene chloride can be used to separate the product, since it less polar than water also. The crude product will prefer to dissolve in the organic, methylene chloride layer, instead of the aqueous portion, therefore separating the two layers. This allows for the aqueous layer to be extracted, and the remaining organic layer, containing the product, is collected. Because the methylene chloride is more dense than water, it sinks to the bottom of the separatory funnel and can be collected. We have seen in Chem 244 Lab, as well as in lecture 21 that methylene chloride, an alkyl halide, is less polar than water and would be able to extract the nonpolar product.
When N-(tert) butoxycarbonyl)L-methionine to HCl, the Boc group will first form a t-butyl carbocation in an elimination reaction. Then the carbon dioxide will be released, and a free amine will form. This allows for the newly-formed diketopiperazine to collect in the methylene chloride layer. The aqueous layer can then be removed to extract the product, which has dissolved in the methylene chloride. The amine product in HCl forms NH3+. Because it is a salt, it remains in the aqueous layer. The crude product however is not a salt, and is neutral, so it stays in the methylene chloride layer.
Why were methylene chloride and hydrochloric acid used to obtain the product?
From “Exp 006
2:30pm (04/02/2006) Extraction: Added 5ml of 5% HCl and 5 ml methylenechloride to the round bottom flask containg the crude product and transferred it to a separatory funnel. Separated the bottom methylenechloride layer from the top aqueous layer. Washed the aquoues acid layer twice with 5mls of methylenechloride and collected the bottom methylenechloride layer each time. Combined the organic layers and dried it over anhydrous magnesium sulfate (MgSO4). Evaporated the dichloromethane using a rotovap and collected the extracted product. A TLC of the product was obtained (Shown above).”
Because the product is soluble in methylene chloride , the methylene chloride can be used to separate the product, since it less polar than water also. The crude product will prefer to dissolve in the organic, methylene chloride layer, instead of the aqueous portion, therefore separating the two layers. This allows for the aqueous layer to be extracted, and the remaining organic layer, containing the product, is collected. Because the methylene chloride is more dense than water, it sinks to the bottom of the separatory funnel and can be collected. We have seen in Chem 244 Lab, as well as in lecture 21 that methylene chloride, an alkyl halide, is less polar than water and would be able to extract the nonpolar product.
When N-(tert) butoxycarbonyl)L-methionine to HCl, the Boc group will first form a t-butyl carbocation in an elimination reaction. Then the carbon dioxide will be released, and a free amine will form. This allows for the newly-formed diketopiperazine to collect in the methylene chloride layer. The aqueous layer can then be removed to extract the product, which has dissolved in the methylene chloride. The amine product in HCl forms NH3+. Because it is a salt, it remains in the aqueous layer. The crude product however is not a salt, and is neutral, so it stays in the methylene chloride layer.
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